A 600 gallon brine tank is to be cleared by piping in pure water at 1 gal/min. , and allowing the well-stirred solution to flow out at the rate of 2 gal/min. If the tank initially contains 1500 pounds of salt, a) how much salt is left in the tank after 1 hour? b) after 9 hours and 59 min?

Volume at any time

$V = 600 + (1 - 2)t$

$V = 600 - t$

Let

Q= amount of salt (in lbs) in the tank at any time.$\dfrac{Q}{600 - t}$ = concentration of salt (in lbs/gal) at any time.

$\dfrac{dQ}{dt} = 0 - \dfrac{Q}{600 - t}(2)$

$\displaystyle \int \dfrac{dQ}{Q} = 2 \int \dfrac{-dt}{600 - t}$

$\ln Q = 2 \ln (600 - t) + \ln C$

$\ln Q = \ln (600 - t)^2 + \ln C$

$\ln Q = \ln C(600 - t)^2$

$Q = C(600 - t)^2$

When

t= 0,Q= 1500 lbs$1500 = C(600 - 0)^2$

$C = \frac{1}{240}$

Hence,

$Q = \frac{1}{240}(600 - t)^2$

When

t= 1 hr = 60 min$Q = \frac{1}{240}(600 - 60)^2 = 1215 ~ \text{lbs}$

When

t= 9 hrs and 59 min = 599 min$Q = \frac{1}{240}(600 - 599)^2 = \frac{1}{240} ~ \text{lbs}$

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